Question

Balance each of the following equations according to the half- reaction method:

(a) Sn2+ + Cu2+ --> Sn4+ + Cu+
(b) H2S + Hg2 2+ --> Hg + S (in acid)
(c) CN- + ClO2 --> CNO- + Cl- (in acid)
(d) Fe2+ + Ce 4+ --> Fe3+ +Ce3+
(e) HBrO --> Br- + O2 (in acid)

Answer 1

Answer : The balanced chemical equation in a acidic solution are,

(a)
`Sn^(2+)+2Cu^(2+)\rightarrow Sn^(4+)+2Cu^+`

(b)
`H_2S+Hg_2^(2+)\rightarrow 2Hg+S+2H^+`

(c)
`5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+`

(d)
`Fe^(2+)+Ce^(4+)\rightarrow Fe^(3+)+Ce^(3+)`

(e)
`2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+`

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is,

`Sn^(2+)+Cu^(2+)\rightarrow Sn^(4+)+Cu^+`

The oxidation-reduction half reaction will be :

Oxidation :
`Sn^(2+)\rightarrow Sn^(4+)+2e^-`

Reduction :
`Cu^(2+)+1e^-\rightarrow Cu^+`

In order to balance the electrons, we multiply the reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation will be,

`Sn^(2+)+2Cu^(2+)\rightarrow Sn^(4+)+2Cu^+`

(b) The given chemical reaction is,

`H_2S+Hg_2^(2+)\rightarrow Hg+S`

The oxidation-reduction half reaction will be :

Oxidation :
`H_2S\rightarrow S+2H^++2e^-`

Reduction :
`Hg_2^(2+)+2e^-\rightarrow 2Hg`

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

`H_2S+Hg_2^(2+)\rightarrow 2Hg+S+2H^+`

(c) The given chemical reaction is,

`CN^-+ClO_2\rightarrow CNO^-+Cl^-`

The oxidation-reduction half reaction will be :

Oxidation :
`CN^-+H_2O\rightarrow CNO^-+2H^++2e^-`

Reduction :
`ClO_2+4H^++5e^-\rightarrow Cl^-+2H_2O`

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

`5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+`

(d) The given chemical reaction is,

`Fe^(2+)+Ce^(4+)\rightarrow Fe^(3+)+Ce^(3+)`

The oxidation-reduction half reaction will be :

Oxidation :
`Fe^(2+)\rightarrow Fe^(3+)+1e^-`

Reduction :
`Ce^(4+)+1e^-\rightarrow Ce^(3+)`

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation will be,

`Fe^(2+)+Ce^(4+)\rightarrow Fe^(3+)+Ce^(3+)`

(e) The given chemical reaction is,

`HBrO\rightarrow Br^-+O_2`

The oxidation-reduction half reaction will be :

Oxidation :
`HBrO+H_2O\rightarrow O_2+Br+3H^++3e^-`

Reduction :
`HBrO+H^++2e^-\rightarrow Br^-+H_2O`

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

`2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+`